# How can you use trigonometric functions to simplify  15 e^( ( 2 pi)/3 i )  into a non-exponential complex number?

Nov 19, 2017

The answer is $= \frac{15}{2} \left(- 1 + i \sqrt{3}\right)$

#### Explanation:

Apply Euler's formula

${e}^{i \theta} = \cos \theta + i \sin \theta$

Therefore,

$15 {e}^{\frac{2}{3} \pi i} = 15 \left(\cos \left(\frac{2}{3} \pi\right) + i \sin \left(\frac{2}{3} \pi\right)\right)$

$\cos \left(\frac{2}{3} \pi\right) = - \frac{1}{2}$

$\sin \left(\frac{2}{3} \pi\right) = \frac{\sqrt{3}}{2}$

So,

$15 {e}^{\frac{2}{3} \pi i} = 15 \left(- \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)$

$= \frac{15}{2} \left(- 1 + i \sqrt{3}\right)$